1 thought on “Find the nature of roots of quadratic equation 3x×x-4 3x + 4 = 0. If real<br />roots exist, lind them.”
Answer:
For a quadratic equation ax² + bx + c =0, the term b² – 4ac is called discriminant (D) of the quadratic equation because it determines whether the quadratic equation has real roots or not ( nature of roots).
D= b² – 4ac
So a quadratic equation ax² + bx + c =0, has
i) Two distinct real roots, if b² – 4ac >0 , then x= -b/2a + √D/2a &x= -b/2a – √D/2a
ii) Two equal real roots, if b² – 4ac = 0 , then x= -b/2a or -b/2a
iii) No real roots, if b² – 4ac <0
SOLUTION:
Given: 3x² – 4√3x + 4 = 0
On Comparing it with ax² + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant(D) = b² – 4ac
D= (-4√3)² – 4(3)(4)
D= 16 × 3 – 48
48 – 48 = 0
As , b² – 4ac = 0,
Hence,the given quadratic equation has real and equal roots.
Answer:
For a quadratic equation ax² + bx + c =0, the term b² – 4ac is called discriminant (D) of the quadratic equation because it determines whether the quadratic equation has real roots or not ( nature of roots).
D= b² – 4ac
So a quadratic equation ax² + bx + c =0, has
i) Two distinct real roots, if b² – 4ac >0 , then x= -b/2a + √D/2a &x= -b/2a – √D/2a
ii) Two equal real roots, if b² – 4ac = 0 , then x= -b/2a or -b/2a
iii) No real roots, if b² – 4ac <0
SOLUTION:
Given: 3x² – 4√3x + 4 = 0
On Comparing it with ax² + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant(D) = b² – 4ac
D= (-4√3)² – 4(3)(4)
D= 16 × 3 – 48
48 – 48 = 0
As , b² – 4ac = 0,
Hence,the given quadratic equation has real and equal roots.
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